Answer to Question #156291 in Linear Algebra for Tony

Question

Let A = {(x, y, z) E R^3ㅣx - y + z = 0} where E represents "an element of" and R the set of real numbers
A mapping f is defined in R^(2*2) , the set of all square matrices of order 2 with entries in R by f: R^(2*2) → R^2
A → Av where v = (-1 2).
(a) Determine Im F. Give a basis of it and precise its dimension.

Answer 1

Let $f: \mathbb R^{2*2} → \mathbb R^2$, $A → Av$ where $v = (-1, 2)$ . Let us show that $Im f=\mathbb R^2$, that is, $f$ is surjective. Fix any $(a,b)\in\mathbb R^2$. Then for the matrix $A=\left(\begin{array}{cc} -a & 0 \\ -b & 0\end{array}\right)$ we have that $Av=\left(\begin{array}{cc} -a & 0 \\ -b & 0\end{array}\right) \left(\begin{array}{cc} -1 \\ 2\end{array}\right)= \left(\begin{array}{cc} a \\ b\end{array}\right)$, and therefore, $Im f=\mathbb R^2$. The standard basis for $Im f=\mathbb R^2$ consists of the linearly independent vectors $e_1=(1,0)$ and $e_2=(0,1)$. Thus, $\dim(Im f)=\dim(\mathbb R^2)=2.$